(a) An electric dipole consists of two point charges q and $-q$ separated by a distance $2a$ . Derive an expression for the electric field E due to this dipole at a point distant r from the centre of the dipole on the equatorial plane. Write the expression for the electric field at a far off point, i.e. $r \gg a$ .

(b) A dipole is placed in x-y plane such that charges q and $-q$ are located at $x = a$ and $x = b$ respectively. There exists an electric field $E = 2\hat{i}\text{ N/C}$ in the region. Calculate the force F and torque τ experienced by the dipole.

OR

(a) Two cells of emf $E_1$ and $E_2$ with internal resistances $r_1$ and $r_2$ respectively, are connected in parallel by connecting their positive terminals together and negative terminals together. Deduce an expression for equivalent emf and equivalent internal resistance of the combination.

(b) A parallel combination, as stated in (a) above, of two cells of emfs $E$ and $3E$ and internal resistances R each is connected across a resistance $2R$ . Find the current that flows through resistance $2R$ .

Q: Why is the torque zero in part (b) even though there is a field?

Torque on a dipole is $\tau = p \times E = pE \sin\theta$, where θ is the angle between p and E. In part (b) both charges lie on the x-axis, so the dipole moment points along î, exactly the same direction as $E = 2\hat{i}$. With $\theta = 0$, $\sin\theta = 0$, so the torque is zero. The force is zero too because the field is uniform.

Q: How does the equatorial field compare with the axial field of the same dipole?

For a far-off point ($r \gg a$), the axial field is $E_{axial} = 2kp/r^3$ and the equatorial field is $E_{equatorial} = kp/r^3$. So the equatorial field is exactly half the axial field in magnitude and points in the opposite direction relative to p.

Electric Charges and Fields

Electric Dipole — Equatorial Field, Force and Torque

Official Answer

This 5-mark answer covers the equatorial dipole field derivation and the force/torque on a dipole in a uniform field.

Part (a) — Field on the equatorial plane

Consider a dipole with charges $+q$ and $-q$ separated by $2a$ , dipole moment $p = q(2a)$ . Take a point P on the equatorial line, at distance r from the centre O.

Distance of P from each charge: $\sqrt{r^2 + a^2}$ .
Field magnitude due to each charge: $E_+ = E_- = q / [4\pi\varepsilon_0 (r^2 + a^2)]$ .
The components perpendicular to the axis (along the equatorial line) cancel; the components parallel to the axis (pointing from +q to −q, i.e. anti-parallel to p) add.

Resultant field:

$E = 2 \cdot E_+ \cdot \cos\theta$ , where $\cos\theta = a / \sqrt{r^2 + a^2}$

$E = (1 / 4\pi\varepsilon_0) \cdot p / (r^2 + a^2)^{3/2}$ , directed anti-parallel to p.

Far-off point ( $r \gg a$ ): neglect $a^2$ against $r^2$ :

$E \approx (1 / 4\pi\varepsilon_0) \cdot p / r^3$ (equatorial field is half the axial field and opposite in direction).

Part (b) — Force and torque

Both charges lie on the x-axis (at $x = a$ and $x = b$ ), so the dipole moment $p = q(a - b)\hat{i}$ is directed along the x-axis. The field $E = 2\hat{i}\text{ N/C}$ is uniform.

Net force: In a uniform field the forces on +q and −q are equal and opposite → $F = 0$ .
Torque: $\tau = p \times E = q(a - b)\hat{i} \times 2\hat{i}$ = 0, because p and E are parallel (angle = 0).

Result: $F = 0$ and $\tau = 0$ .

OR alternative (cells in parallel): Equivalent emf $E_{eq} = (E_1 r_2 + E_2 r_1)/(r_1 + r_2)$ and $r_{eq} = r_1 r_2/(r_1 + r_2)$ . For E and 3E with R each: $E_{eq} = 2E$ , $r_{eq} = R/2$ , so current through 2R = $2E/(R/2 + 2R)$ = $4E/5R$ .

equatorial planedipole momentelectric fielduniform fieldnet force zerotorque1/r^3 dependencesuperposition

Marking Scheme

11 mark: geometry/diagram — point P on equatorial line, distance $\sqrt{r^2+a^2}$ from each charge, perpendicular components cancel.
21.5 marks: derivation $E = (1/4\pi\varepsilon_0)\cdot p/(r^2+a^2)^{3/2}$ directed anti-parallel to p.
30.5 mark: far-field result $E \approx (1/4\pi\varepsilon_0)\cdot p/r^3$ for $r \gg a$ .
41 mark: net force $F = 0$ (uniform field → equal and opposite forces).
51 mark: torque $\tau = p \times E = 0$ because p (along x) is parallel to E (along x).
6OR: 2 marks $E_{eq} = (E_1 r_2+E_2 r_1)/(r_1+r_2)$ ; 1 mark $r_{eq} = r_1 r_2/(r_1+r_2)$ ; 2 marks current = $4E/5R$ .

Hint

Equatorial field: superpose the two charge fields; only components anti-parallel to p survive → $E = kp/(r^2+a^2)^{3/2}$ . In part (b) note both charges lie on the x-axis, so $p \parallel E$ .

Quick Oral Answer

On the equatorial plane the dipole field is $E = kp/(r^2+a^2)^{3/2}$ directed opposite to p, becoming $kp/r^3$ far away. If the dipole moment lies along the field, both the net force and the torque are zero — exactly the case in part (b) where both charges sit on the x-axis.

Analysis & Explanation

Concept. This question links two staples of electrostatics: building the equatorial dipole field by vector superposition, and analysing a dipole in a uniform field where the net force vanishes but a torque (usually) acts.

Why the equatorial field is anti-parallel to p. On the axial line the field points along p; on the equatorial line the surviving components point from +q toward −q, i.e. opposite to p. Its magnitude at far points is exactly half the axial value — a symmetry students should remember ( $E_{axial} = 2kp/r^3$ , $E_{equatorial} = kp/r^3$ ).

The exam trap in part (b). Many students reflexively write $\tau = pE \sin\theta$ and plug in a non-zero value. Here both charges sit on the x-axis, so p is along $\hat{i}$ , exactly parallel to $E = 2\hat{i}$ . With $\theta = 0$ , $\sin\theta = 0$ , giving zero torque; and a uniform field gives zero net force. The whole point is to test whether you recognise the geometry rather than blindly apply a formula.

Real-world link. The $1/r^3$ dipole fall-off (faster than a point charge’s $1/r^2$ ) explains why polar molecules like water exert only short-range electrostatic influence, and it underlies antenna radiation patterns and the behaviour of dielectrics in capacitors.

OR part. Cells in parallel share the load current; the equivalent-emf result $E_{eq} = (E_1 r_2 + E_2 r_1)/(r_1+r_2)$ comes from combining terminal-voltage equations — a common numerical in Current Electricity.

Common Mistakes

1Writing a non-zero torque in part (b): since both charges are on the x-axis, $p$ is along $\hat{i}$ and parallel to E, so $\tau = pE \sin 0^\circ = 0$ .
2Forgetting that on the equatorial line the field is directed anti-parallel to the dipole moment (opposite to the axial case).
3Dropping the $(r^2+a^2)^{3/2}$ denominator too early and using $r^3$ before stating the $r \gg a$ approximation.

Interesting Facts

The equatorial dipole field is exactly one-half the axial field at the same distance — and points the opposite way — a result that follows purely from geometry.

The $1/r^3$ fall-off means a neutral polar molecule’s influence dies far faster than a point charge’s $1/r^2$ , which is why van der Waals dipole forces are short-range.

A uniform field can rotate a dipole but never translate it; net translation needs a non-uniform field, which is how a charged comb attracts neutral paper bits.

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Frequently Asked Questions

Why is the electric field on the equatorial plane opposite to the dipole moment?

On the equatorial line, the horizontal components (perpendicular to the axis) of the two charge fields cancel, while the components along the dipole axis add. These surviving components point from the positive charge toward the negative charge, which is anti-parallel to the dipole moment p (which points from $-q$ to $+q$ ). Hence the equatorial field is directed opposite to p.

Why is the torque zero in part (b) even though there is a field?

Torque on a dipole is $\tau = p \times E = pE \sin\theta$ , where θ is the angle between p and E. In part (b) both charges lie on the x-axis, so the dipole moment points along î, exactly the same direction as $E = 2\hat{i}$ . With $\theta = 0$ , $\sin\theta = 0$ , so the torque is zero. The force is zero too because the field is uniform.

How does the equatorial field compare with the axial field of the same dipole?

For a far-off point ( $r \gg a$ ), the axial field is $E_{axial} = 2kp/r^3$ and the equatorial field is $E_{equatorial} = kp/r^3$ . So the equatorial field is exactly half the axial field in magnitude and points in the opposite direction relative to p.